Dividing a quantity in a given ratio

What to assume they know

• Multi-term ratios, and finding the common factor from one known quantity — Ratios of three or more quantities at once
• Sharing a whole in a two-term ratio — Sharing a whole in a given ratio
• Adding decimals, and dividing a whole number by a decimal such as 5.5
• That the three angles of a triangle add to 180°
• The triangle inequality in the informal form met in earlier classes: two sides together must exceed the third
• Constructing a triangle from three given angles, and from three given sidelengths, with ruler, compasses and protractor

What they should be able to do

• Add the terms of a ratio and interpret the sum as a count of equal parts
• Divide a stated whole by that sum to get one part, and multiply through to get every share
• Write the same division in its fraction form, term over sum of terms, and use it directly
• Show that the shares reconstruct the whole, and explain why they must
• Divide a quantity given in decimals, or a quantity that is not a whole number of parts, without rounding prematurely
• Distinguish a problem where the total is given from one where a single part is given, and choose the right first step for each
• Apply the division to a total that is fixed by mathematics rather than by the question — the 180° of a triangle
• Decide whether a ratio of sides determines a unique triangle, and whether a given ratio of sides can be a triangle at all

Where it usually goes wrong

• "Divide 12 in the ratio 2 : 1 means 12 ÷ 2 and 12 ÷ 1." That gives 6 and 12, which add to 18. The division is by the sum of the terms, and the check is that the shares add back to 12.
• "The sum of the terms is the whole." In Example 3 the terms add to 5.5 and the whole is 110. The sum counts parts; the whole is measured in units. Keeping those two numbers visibly distinct is most of the battle.
• "You can't divide by 5.5." You can, and the chapter does. Sums of ratio terms are not obliged to be whole numbers.
• "288 Odiya books means 288 books in all." Item 2 hands you a part, not the whole. Diagnosing which of the two you have been given is the skill; the arithmetic afterwards is identical.
• "Coins in the ratio 4 : 3 : 2 : 1 means money in the ratio 4 : 3 : 2 : 1." It does not. Ten coins of one rupee and ten of ten rupees are equal in count and a factor of ten apart in value.
• "Same ratio of sides means the same triangle." Same shape, any size. The words similar and congruent do different jobs, and exercise item 4 is built to separate them.
• "Any three numbers can be the sides of a triangle." 1 : 3 : 5 cannot, at any scale. Multiply all three by anything you like and the short two still fail to reach past the long one.
• "A ratio always determines the answer." Only against a fixed total. Sections 8 to 10 are three variations on that one sentence.

Questions to check understanding

• Divide a stated quantity in a stated three- or four-term ratio, showing the sum of the terms and the size of one part
• Divide a quantity where the sum of the terms is not a whole number
• Given one share and the ratio, recover the whole
• Given a part rather than the whole, find the remaining shares — the item type students most often start wrongly
• Split 180° among three angles in a given ratio, and construct the triangle
• Decide whether triangles from a given side ratio must be congruent, with reasons
• Decide whether a given side ratio can be a triangle at all, with reasons
• Convert a count ratio into a value ratio, as the coins item requires

Examples worth working on the board

Values marked printed are worked out on the page. Values marked not in the book are worked out here on the chapter's inputs. The chapter prints answers for its Examples and none at all for its Figure it Out items.

• The two-term recap (Part II §3.4, p.58). Printed: to split 12 in the ratio 2 : 1, add the terms to get 3, divide the whole by that sum to get 4, then multiply each term by 4, giving 8 and 4.
• Example 3, the concrete (Part II §3.4, p.59). Inputs: 110 units of concrete are wanted; cement, sand and gravel go in as 1 : 1.5 : 3. Printed: one unit of cement brings 1.5 of sand and 3 of gravel, which is 5.5 units of concrete together; 110 ÷ 5.5 = 20, so the batch is that group twenty times over, and 1 × 20 = 20 units of cement, 1.5 × 20 = 30 units of sand, 3 × 20 = 60 units of gravel.
• The figure for Example 3 (Part II p.59, artwork). A dark mound labelled 110 units of concrete at the top, with three arrows fanning down from it, labelled 1, 1.5 and 3, each ending on a separate pile: a grey pile labelled 20 units of cement, an orange-brown pile labelled 30 units of sand, and a pile of stones labelled 60 units of gravel. Every one of those labels is artwork lettering and was read off the printed page. The figure is worth redrawing because the arrow labels are the ratio terms while the pile labels are the shares — the same figure holds both, which is exactly the distinction section 6 needs.
• The general fraction rule (Part II p.59). Printed: to divide a quantity x in the ratio a : b : c and so on, each share is x multiplied by that term over the sum of all the terms. The SUMMARY restates the same rule on Part II p.69 with the letters p, q, r, s.
• Why the shares close (not in the book, and the argument of section 5). The fractions 1/5.5, 1.5/5.5 and 3/5.5 add to 5.5/5.5, which is 1. So the three shares add to one whole x, whatever x is and whatever the ratio is. Check it on the numbers: 20 + 30 + 60 = 110. This is the reason the rule is trustworthy rather than merely usable, and the chapter does not state it.
• Example 4, the purple paint (Part II §3.4, p.59). Inputs: red, blue and white in the ratio 2 : 3 : 5, and 50 ml of purple wanted in total. Printed: the sum of the terms is 10, and the three shares come out as 10 ml, 15 ml and 25 ml, each shown as 50 multiplied by that term over 10.
• The same paint, two directions (not in the book). §3.3's Example 1 gave Yasmin 10 litres of white and asked for the rest; §3.4's Example 4 gives 50 ml of finished paint and asks for all three. Same ratio, opposite direction, and in both cases the first thing computed is the size of one part — 2 litres there, 5 ml here. Putting the two side by side is the cleanest way to make the thesis land.
• Example 5, the triangle from its angles (Part II §3.4, p.60). Inputs: build a triangle whose angles stand in the ratio 1 : 3 : 5. Printed: the angle sum is 180°, the terms add to 9, and the three angles are 180 × 1/9 = 20°, 180 × 3/9 = 60° and 180 × 5/9 = 100°. A pale blue scalene triangle is drawn beside the working, lettered A at the left vertex marked 20°, B at the right vertex marked 60°, and C at the top right marked 100°.
• Figure it Out, Part II p.60, five items. All inputs, no printed answers:
  • A coaching session of 150 minutes split across warm-up and cool-down, batting, bowling and fielding in the ratio 3 : 4 : 3 : 5.
  • A library holding Odiya, Hindi and English books in the ratio 3 : 2 : 1, with 288 Odiya books on the shelves; the Hindi and English counts are wanted.
  • A hundred coins made up of ten-rupee, five-rupee, two-rupee and one-rupee coins in the ratio 4 : 3 : 2 : 1; the total money is wanted.
  • Construct a triangle whose sides are in the ratio 3 : 4 : 5, then decide whether every triangle drawn from that ratio is congruent to every other, with reasons. Flagged in the margin as a Math Talk item.
  • Decide whether a triangle can be built with sides in the ratio 1 : 3 : 5, with reasons.
• Worked answers to those five (not in the book — none of this is printed): item 1, the terms add to 15 and 150 ÷ 15 = 10 minutes per part, giving 30, 40, 30 and 50 minutes; item 2 is the odd one out because a part, not the whole, is given — 288 ÷ 3 = 96 books per part, so 192 Hindi and 96 English, and the library's total of those three languages is 576; item 3, 100 ÷ 10 = 10 coins per part, so 40 ten-rupee, 30 five-rupee, 20 two-rupee and 10 one-rupee coins, worth ₹400 + ₹150 + ₹40 + ₹10 = ₹600 — note the coin counts are in 4 : 3 : 2 : 1 while the values are in 400 : 150 : 40 : 10, which is a different ratio entirely, and that is the trap; item 4, the triangles all have the same shape but any size, so they are similar and not in general congruent — the ratio fixes no length; item 5, no triangle exists, because 1 + 3 is less than 5 and two sides must together beat the third.
• The angles-against-sides contrast (not in the book, section 9, and the sharpest point available here). An angle ratio determines a triangle's angles outright because the angle total is nailed down at 180° before the question starts. A side ratio determines nothing about size, because there is no fixed perimeter to divide. The chapter sets Example 5 and exercise items 4 and 5 within half a page of each other and never draws the comparison.

Figures to have open

• The chapter's Example 3 figure (Part II p.59): the concrete mound above three material piles, with the ratio terms on the arrows and the shares on the piles. Redraw it as a schematic. The double labelling is the whole reason to keep this figure, and the printed version's photographic piles hide it.
• A parts-bar: a single bar of the whole, subdivided into the sum of the ratio terms as equal cells, then grouped and coloured by term. This one figure carries sections 2 to 6 and should be reused rather than redrawn each time.
• The chapter's Example 5 triangle (Part II p.60), lettered A, B, C with 20°, 60° and 100° at the matching vertices. Standard schematic.
• Three nested triangles with sides in 3 : 4 : 5 at three different sizes, for section 9. Standard schematic; the textbook does not supply this and section 9 cannot be made without it.
• Two short segments of lengths 1 and 3 laid against a segment of length 5. Standard schematic.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part II, printed Chapter 3, "Proportional Reasoning-2", §3.4 "Dividing a Whole in a Given Ratio", Part II pp.58–60. §3.4 begins near the foot of Part II p.58; Example 3 and the general fraction rule and Example 4 are on Part II p.59; Example 5 and the five-item Figure it Out are at the top of Part II p.60, above the start of §3.5.
• Part II p.69, the chapter SUMMARY, second bullet, for the fraction rule in the letters p, q, r, s.
• Backward pointers: Ratios of three or more quantities at once for the same concrete and paint ratios read in the scaling direction; Part I printed Chapter 7, §7.5, for sharing a whole in a two-term ratio.
• Forward pointer inside this chapter: §3.5 begins in the lower half of Part II p.60 and applies this section's rule to a whole of 360°; it is covered by Building a pie chart: turning counts into angles.